Show finite complement topology is, in fact, a topology

My attempt to prove the following is below:

Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$

My book calls this set the “finite complement topology”

I just want to know if this proof is correct. (I intuitively understand why its a topology, but I’m not sure if I’m showing it properly)

Using the definition of topology:

$\mathscr{T}$ is a topology on X if and only if the following are true:

(i) X and Ø are elements of $\mathscr{T}$.

(ii) $\mathscr{T}$ is closed under finite intersections.

(iii) $\mathscr{T}$ is closed under arbitrary unions.

Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2…$ are all finite sets)

So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.

Now to verify the definition.

(i) X and Ø are elements of $\mathscr{T}$.

Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.

(ii) $\mathscr{T}$ is closed under finite intersections.

$U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan’s law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| < |\mathbb{N}| $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.

(iii) $\mathscr{T}$ is closed under arbitrary unions.

$U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan’s law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| < |\mathbb{N}| $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.
Now let’s consider the infinite case: $V_k \cap … \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap … $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.

Solutions Collecting From Web of "Show finite complement topology is, in fact, a topology"

(i) is fine, (ii) is fine too, although you can do the finite case in one go (not go via 2), as “De Morgan” holds for any family of sets. The $U_i, V_i$ notation is a bit weird, and you need to consider the special case set as well:

Let $U_1,\ldots U_n$ be open in the topology. We want to show the intersection is open, so if any of them is empty, the intersection will be too, so we can discard those cases: assume all $U_i$ are non-empty and open, so their complements are finite. Now De Morgan says:

$$X \setminus (\cap_{i=1}^n U_i ) = \cup_{i=1}^n (X \setminus U_i)$$

and so the right hand side is finite, as a finite union of finite sets, and so $\cap_{i=1}^n U_i$ is cofinite, and hence open.

Now unions, which is more straightforward than you seem to think (no proof by contradiction is needed):

Suppose $U_i, i \in I$ are open. We want to see that $\cup_i U_i$ is open, so we can discard all $U_i$ that happen to be empty, because they do not affect the union. So assume all $U_i$ are cofinite, and we can assume we have at least one (so $I$ is non-empty, or we have an empty union, which is open, so OK).

Now de Morgan again:

$$X \setminus (\cup_i U_i) = \cap_i (X \setminus U_i)$$

and we see that the right hand set is an intersection of at least one finite set (and the intersection only can become smaller with more sets) so the right hand side is surely finite, and $\cup_i U_i$ is cofinite, and we’re done.

To show that an arbitrary union is in the topology, it is simpler to tackle every case directly:

Let $A=\bigcup_{i \in I} U_i$ be an arbitrary union of sets whose complements are are finite. Then by De Morgan’s law, $A=X-(\bigcap_{i \in I}V_i)$, for some finite sets $V_i$. Now by the definition of intersection, $\bigcap_{i \in I}V_i \subseteq V_i$ for all $i \in I$. But since $V_i$ is finite, so is any one of its subsets. Hence the complement of $A$ is finite, implying that $A$ is an open set.

Otherwise your arguments look good.

The problem is that, the set of finite subsets of $X$ may not be countable. So, saying $V_1, V_2, \dots$ are finite subsets is a little bit problematic.

For $(ii)$, you can simply say that:

Let $U_1, U_2 \in \mathscr T$. Let $U = U_1 \cap U_2$. Then $U^c = U_1^c \cup U_2^c$ is a union of two finite sets. Hence $U^c$ is finite and thus $U \in \mathscr T$.

For $(iii)$,

$Let \{U_\alpha \}$ be a set of opens in $\mathscr T$ (with $U_\alpha \neq \emptyset$). Let $U = \bigcup U_\alpha$. Then, $U^c = \bigcap U_\alpha^c$ is an intersection of finite sets, which is also finite. Hence $U^c$ is finite. Therefore $U \in \mathscr T$.